8.4 Integration as the limit of a sum — Practise with help

1. $3 + 5 + 7 + 9 + 11 = 35$.
2. $\dfrac{10 \times 11}{2} = 55$.
3. $\left[\dfrac{x^2}{2}\right]_0^2 = 2$.

If question 2 was hard, revise sigma notation in Section 4.4.

The width is the length of the interval divided by $n$. The right end of strip $i$ is $i$ widths from the left.

The interval has length $4$, so $\delta x = \dfrac{4}{n}$ and $x_i = i\delta x = \dfrac{4i}{n}$.

The height of strip $i$ is $x_i^2 = \dfrac{16i^2}{n^2}$, so the sum is $\displaystyle\sum_{i=1}^{n} \frac{16i^2}{n^2} \times \frac{4}{n} = \frac{64}{n^3}\sum_{i=1}^{n} i^2$.

With $n$ strips, $\delta x = \dfrac{1}{n}$ and $x_i = \dfrac{i}{n}$. Use $\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.

$S_n = \displaystyle\sum_{i=1}^{n} \frac{i}{n} \times \frac{1}{n} = \frac{1}{n^2}\sum_{i=1}^{n} i = \frac{1}{n^2} \times \frac{n(n+1)}{2}$.

As $n \to \infty$, $\dfrac{n+1}{n} \to 1$, so $S_n \to \dfrac{1}{2}$. Therefore $\displaystyle\int_0^1 xdx = \frac{1}{2}$.

Here $\delta x = \dfrac{2}{n}$ and $x_i = \dfrac{2i}{n}$. Use $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$.

$S_n = \displaystyle\sum_{i=1}^{n} \frac{4i^2}{n^2} \times \frac{2}{n} = \frac{8}{n^3} \times \frac{n(n+1)(2n+1)}{6}$. Simplify to a fraction in $n$.

Expand the top: $(n+1)(2n+1) = 2n^2 + 3n + 1$. Divide by $n^2$ and let $n \to \infty$:

So $\displaystyle\int_0^2 x^2dx = \frac{8}{3}$.

Split the sum into two parts: one with $i$ and one with the constant $1$. The constant part is $\displaystyle\sum_{i=1}^{n} 1 = n$.

$\delta x = \dfrac{3}{n}$ and $x_i = \dfrac{3i}{n}$, so $S_n = \displaystyle\sum_{i=1}^{n} \left(\frac{3i}{n} + 1\right)\frac{3}{n} = \frac{9}{n^2}\sum_{i=1}^{n} i + \frac{3}{n}\sum_{i=1}^{n} 1$.

As $n \to \infty$, the first fraction tends to $\dfrac{9}{2}$, so

Therefore $\displaystyle\int_0^3 (x + 1)dx = \frac{15}{2}$.

Compare the left and right heights of one strip. How large is their difference, and what happens to it as the strips get thinner?

For a strip of width $\delta x$, the height changes by about $f'(x)\delta x$ across the strip. That is the difference between the left and right heights.

Across one strip the height changes by roughly $f'(x)\delta x$. The area of the strip is therefore altered by at most about $f'(x)(\delta x)^2$ when the endpoint is moved from one end to the other. There are $n$ strips and $\delta x = \dfrac{b-a}{n}$, so the total change is at most about

As $n \to \infty$ this tends to $0$, so the left and right sums have the same limit. The choice of endpoint does not affect the integral.

Integrate each part as a separate sum. The $x^2$ part uses the sum of squares and the $1$ part uses $\displaystyle\sum_{i=1}^{n} 1 = n$.

$\delta x = \dfrac{2}{n}$ and $x_i = \dfrac{2i}{n}$, so $S_n = \dfrac{8}{n^3}\sum i^2 + \dfrac{2}{n}\sum 1$.

The first fraction tends to $\dfrac{8}{3}$, so

The Fundamental Theorem of Calculus gives the same value: $\left[\dfrac{x^3}{3} + x\right]_0^2 = \dfrac{8}{3} + 2$.